How can I understand that an atom subjected to an oscillating electric field (e.g., $\vec{E}=\hat{i} E_0\cos\omega t$) behaves like an oscillating electric dipole? What is the underlying picture that comes out of the quantum mechanical description of the atom?

  • $\begingroup$ What do you mean when you say "oscillating dipole"? Oscillating with respect to what? $\endgroup$ – Aaron Stevens Jul 26 '18 at 21:18

Well, let’s start with a static electric field: the electrons would move in the opposite direction of it and the nucleus in the same direction. But the nucleus’ and the electrons’ attractive force counteracts this process, thus forming a (steady) dipole. I know, the picture I painted is very classical, but that’s sufficient. To solve it quantum mechanically you would have to use perturbation theory, only to find the same result. Now if the external electric field oscillates, the dipole will oscillate accordingly.

  • $\begingroup$ You actually don't need possitive and negative charges to move oppositely. A single electron, for the fact of displacing from the origin, stops being a point charge at the origin, so a dipolar term arises. It's just about the origin you use to describe the motion. $\endgroup$ – FGSUZ Jul 26 '18 at 23:13
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    $\begingroup$ The picture you painted is indeed very classical, and unfortunately it is not sufficient. The nucleus mass is much heavier than the electron, and thus their response to the oscillating field is not the same. So your claim that "if the external electric field oscillates, the dipole [which you define as coming from nucleus and electron separation] will oscillate accordingly" is not accurate. In fact, you typically assume the nucleus is frozen during electron dynamics, in the Born-Oppenheimer sense. $\endgroup$ – wcc Jul 26 '18 at 23:37
  • $\begingroup$ This answer seems to be insufficient and overly simplistic with regard to "perturbation theory." What type of perturbation theory would you use? How exactly would you solve the problem for a spatially uniform field, as indicated by the OP (spatially uniform but time dependent)? $\endgroup$ – hft Jul 27 '18 at 5:10
  • $\begingroup$ @IamAStudent depends on your definition of the word “accordingly”. But yes, of course the nucleus’ displacement is much less. $\endgroup$ – Antaios Jul 28 '18 at 16:28
  • $\begingroup$ @hft he changed his question, I didn’t know he wanted a quantum mechanical explanation. $\endgroup$ – Antaios Jul 28 '18 at 16:30

Let's first clarify the coordinate system. We can regard the atom as an effectively two-body system, consisting of the nucleus and the electron(s). The nucleus is very heavy compared to the electrons, so they are effectively infinitely heavy. Thus, we take out the center-of-mass motion and just consider the internal motion of electrons around the nucleus. So the origin is assumed to be fixed at the center of the atom.

In the non-relativistic limit, the electron Hamiltonian with the external EM field is

$H = \frac{(p-qA_{\perp})^2}{2m} + (-g_{q}\frac{q}{2m} S\cdot B)+ V_{Coulomb}(r)$

where $A_{\perp}$ is the transverse part of the vector potential, $g_e$ is the charged particle g-factor, $S$ is the spin and $B$ is the magnetic field. $V_{Coulomb}$ is the Coulomb potential felt by the electron.

$V_{Coulomb}$ gives rise to atomic level structure. $S\cdot B$ is the Zeeman interaction. The dipole interaction between the electron and the external field comes from $p \cdot A_{\perp}$ in the first term of the Hamiltonian. (The $A^2$ term does not couple to the electron obviously). To make this term more physically clear, you can make a unitary transformation so that you get rid of the $A_{\perp}$ and the dipole interaction is written as $-d\cdot E_{\perp}$, where $d = q\cdot r$ is the dipole operator. To be exact, this works as long as the external field does not vary over the extent of the electron wavefunction (long-wavelength approximation).

So where does the oscillating dipole moment come from? First, a non-zero external field in $-d \cdot E_{\perp}$ couples the ground state of the electron to excited state(s): you get the matrix elements $\sum_{e} \langle e | d | g \rangle$. In first-order perturbation theory, the ground state acquires an admixture of the excited states. So the expectation value of the dipole operator is non-zero after the first-order correction to the ground state wavefunction. The external field has induced the dipole moment. When the external field is sinusoidal, the dipole moment, or the expectation value of the dipole operator, follows the drive frequency, as you can derive using time-dependent first-order perturbation theory. This picture of the dipole moment using perturbation theory is standard in atomic physics.

The distinguishing of the transverse and the longitudinal part of the external field is about separating electric field radiation from electric field by static charges. The gauge choice ($\nabla \cdot A = 0$) is called the Coulomb gauge and implies radiation is all contained in the transverse part. See "Photons and Atoms" by Cohen-Tannoudji for a more thorough explanation.

  • $\begingroup$ To add emphasis, the induced dipole moment of an electron by external field has to do with how it couples to excited states. If there were an hypothetical atom with all excitated states having the same parity as the ground state, there is no induced dipole moment, because the dipole operator flips parity. $\endgroup$ – wcc Jul 27 '18 at 0:33
  • $\begingroup$ The OP asks about a time-dependent field. This answer does not appear to address that aspect of the question. $\endgroup$ – hft Jul 27 '18 at 5:12
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    $\begingroup$ I added a short sentence saying about a sinusoidal field. The extension from static to time-dependent case is not important (just use time-dependent theory, at the same level of perturbation order), what is really important is the matrix element between ground and excited states. $\endgroup$ – wcc Jul 27 '18 at 5:14
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    $\begingroup$ Even if the amplitude of the light is spatially independent, as long as it is non-zero, the dipole interaction term gives rise to a non-zero matrix element: $-E_0 \langle e | \hat{d} |g \rangle $ (I wrote $E_0$ as a number, because I'm treating the light classically). $\endgroup$ – wcc Jul 27 '18 at 5:26
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    $\begingroup$ You are correct. I was mistaken, I thought OP asked about a perturbation that only depends on time, but actually he asked about a electric field that only depends on time. In this case the perturbation hamiltonian is the dipole hamiltonian ~$\vec E \cdot \vec x$, which depends on space and time and can cause transitions... that's what I get for staying up too late on stackexchange... sorry about that. $\endgroup$ – hft Jul 27 '18 at 18:22

How can I understand that an atom subjected to an oscillating electric field $\vec{E}=\hat{i} E_0\cos\omega t$ [...]?

The notation appears to be that of a spatially independent and time-dependent field. There could be spatial dependence hidden in the $E_0$, but I will assume it is constant.

Suppose, for simplicity, that the unperturbed Hamiltonian ($H_0$) is that of a single-electron atom. Call the eigenstates of this Hamiltonian $|n>$.

In this case we can treat the perturbation Hamiltonian ($V$) as: $$ exE_0cos(\omega t)\;, $$ where $e$ is the magnitude of the electron charge. (To understand this, remember that the energy is written in terms of the electric potential as $q\phi$, where $q$ is the charge $-e$. And also note that the electric potential $\phi$ is $-xE_0cos(\omega t)$, which was chosen such that $-\nabla \phi=\hat{i}E_0cos(\omega t)$.) I'm also assuming that the $\hat{i}$ is a unit vector in the x direction.

The problem is probably easiest to solve quantum mechanically in the interaction picture (See, for example, Sakurai "Modern Quantum Mechanics" at Section 5.5).

If we are interested in the dipole moment, we need to take the expectation of the dipole operator with respect to the full Hamiltonian eigenstates $|{\Psi_n}>$. The x-component of the dipole operator is $qx.$ So, we are interested in the matrix elements: $$ d(t)=q<x>=-e<\Psi_0 (t)|x|\Psi_0 (t)>=-e<\Psi_0|_Ix_I(t)|\Psi_0 (t)>_I\;, $$ where the subscript "I" refers to the interaction picture, and un-subscript operators and states are in the Schrodinger picture.

In the interaction picture (see Sakurai Section 5.6), we can use time-dependent perturbation theory to see that: $$ |\Psi_0>_I = |0> + \sum_n c_n^{(1)}|n> + O(E_0^2)\;, $$ where $|0>$ is the state prior to turning on the interaction at $t=t_0$ and where the $O(E_0)$ term represents higher-order corrections.

Assuming that the unperturbed ground state does not have a permanent dipole moment, the zeroth order term vanishes. The first order term is: $$ -2e\sum_n Re(<0|x|n>e^{i(E_0-E_n)t/\hbar}c_n^{(1)})=-2e\sum_n Re(<0|x_I(t)|n>c_n^{(1)})\;, $$ where $Re()$ means the real part of the complex expression.

The coefficient $c_n^{(1)}$ is given in Sakurai Eq. 5.6.17 as: $$ c_n^{(1)}=\frac{-i}{\hbar}\int_{t_0}^{t}dt'<n|V_I(t')|0>\;. $$

Now, remember that $V_I$ is given by: $$ ex_I(t)E_0cos(\omega t)\;, $$ where the only operator (based on our assumption of constant field) is the explicit x.

Thus we find that the value of the dipole is: $$ d(t)=-2e^2 E_0\sum_n Re(\frac{-i}{\hbar}\int_{t_0}^t dt'<0|x_I(t)|n><n|x_I(t')|0>cos(\omega t'))\;, $$

Well, that is a nice looking expression, isn't it? We got one factor of x from the interaction and one factor of x from the definition of the dipole operator. We can actually clean this up a little more by performing the sum over $n$, since $$ 1=\sum_n |n><n|\;, $$ to find that $$ d(t) =-2e^2 E_0 Re(\frac{-i}{\hbar}\int_{t_0}^t dt'\chi(t,t')\cos(\omega t'))\;, $$ where we have now written the dipole moment in terms of the time-dependent ground-state position-position correlation function, which I defined as $\chi(t,t')\equiv <0|x_I(t)x_I(t')|0>$.

Of course, we could also go the other way of being less clean and more explicit and re-write the whole thing as: $$ d(t)=-2e^2 E_0\sum_n |<0|x|n>|^2 \frac{1}{\hbar}\int_{t_0}^t dt' \sin((E_0-E_n)(t-t')/\hbar)cos(\omega t')\;. $$

At this point we can't go much further without explicitly defining the unperturbed system so that we can actually start calculating the x matrix elements.


From MIT OCW 8.06 Ch.2 Lecture Notes

Suppose that we have charges $q_1,\dotsc,q_N$ at positions $\vec{x}^{(1)},\dotsc,\vec{x}^{(N)}$, and we apply an electric field $\vec{E}(\vec{x})$. The energy is determined by the scalar potential $\phi(x)$ which is related to the electric field by $\vec{E}=-\vec{\nabla}\phi$. If $\vec{E}(\vec{x}) = \vec{E}$ (i.e. independent of $\vec{x}$) then one possible solution is $\phi(\vec{x})=-\vec{x} \cdot \vec{E}$. In this case the Hamiltonian will be: $$ H = \sum_{i=1}^N q_i\phi(\vec{x}^{(i)}) = \sum_{i=1}^{N}q_i\vec{x}^{(i)}\cdot\vec{E} = -\vec{d}\cdot\vec{E}. $$ where we have defined the dipole moment $\vec{d} = \sum_{i=1}^{N}q_i\vec{x}^{(i)}$.

So it appears what you call a dipole may merely be a definition question. This is for a general case of a collection of particles acting in an $\vec{E}$ field. We only ask that it be independent of position, so this satisfies your criteria. If it is spatially dependent, that can be treated as a perturbation to some extent.

To address the oscillation, it appears that an eigenstate of the Hamiltonian will not be an eigenstate of position (thus "oscillating").

Unit wise, the dipole moment is charge * distance, and so as long as we can define both a distance (in this case an arbitrary displacement $\vec{x}$) and charge (the charge on the particle). For the case of an atom there are several particles (e.g. $N$ of them as in the above example).

I hope the formatting is clear, am new to this.

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    $\begingroup$ I don't understand your notation of $\vec x * \vec E$ What do you mean by this? Do you mean to say the the field is pointing in the x-direction, so then $\phi (\vec x) = -xE$, where $x$ is the x-coordinate of the position vector $\vec x$, and $E$ is the magnitude of $\vec E$? (i.e. $\vec E=E \hat x$, with $\hat x$ being the unit vector in the x-direction, not the unit vector in the direction of the position vector $\vec x$. Maybe I should have used $\vec r$ for $\vec x$ haha) $\endgroup$ – Aaron Stevens Jul 26 '18 at 21:54
  • $\begingroup$ Yes it should be a dot product between the position vector $\vec{x}$ and the electric field $\vec{E}$. I used $\vec{x}$ as that was the notation in the source text. $\endgroup$ – OctopusHelicopter Jul 26 '18 at 22:54
  • $\begingroup$ This answer does not address the question. The field in the OP's question is explicitly time-dependent. $\endgroup$ – hft Jul 27 '18 at 5:12
  • $\begingroup$ This $\vec{E}$ field is arbitray, e.g $\vec{E} = E_0\cos{(\omega t - kz)}$ $\endgroup$ – OctopusHelicopter Jul 27 '18 at 15:40

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